There are functions of the form erx that solve the...

There are functions of the form $e^{rx}$ that solve the differential equation Identify a fundamental set of solutions for this differential equation. $\{e^{-2x}, e^{4x}\}$ $\{e^{-4x}, e^{-2x}\}$ $\{xe^{-2x}, e^{-2x}\}$ $\{e^{-4x}, e^{2x}\}$ $\{e^{2x}, e^{4x}\}$ $y'' + 6y' + 8y = 0$

Transcript

00:03 Show that every function have defined by given that function f is equal to c1 e to the power 2x plus c2 exponential to the power minus 4x.

00:17 Where c1 and c2 are arbitrary constant is a solution of differential equation and the differential equation is given as d square y over d x square minus 2, dy over d x x.

00:43 Minus 8y is equal to 0 so let me check once again we need to check that fx is a solution of this differential equation so we can do two things here is to substitute the value of fx as a y in differential equation and other thing we can do just solve that differential equation and check that whether we get the same solution or note okay because if this gives the same solution then it is automatically prove that this is a solution of differential equation other thing let me use the second method okay first method is easy you just need to take y is equal to c1 exponential 4x plus c2 exponential minus 2x and then substitute those values in equation and check that whether those satisfy they automatically is that is why this one is the solution of equation so let me use the second method so this is the differential equation so we can only need to evaluate the complementary function that complementary function with will be the solution of this equation so let capital d is equal to d over d x if capital d is equal to d over d x it means uh d d d square is d2 over d x square so let me substitute this value in equation we get d square minus 2d minus 8 into y is equal to 0 so if this is equal to 0 why cannot be equal to 0 so it means d square minus 2d minus 8 is equal to 0 so if we solve this equation this is normally differential equation so that will be d square minus 4d plus 2d minus 8 is equal to 0 so that will give us d minus 4 and d plus 2 so if these there are two values then the we get d is equal to 4 and d is equal to minus 2 so the solution of this differential equation will be equal to we get by is equal to c1 which function is first we use that exponential i think there is something wrong with it minus 4 plus no i think that is c1 exponential minus 2x sorry so that will be 4 x and this is minus 2 so 4 x minus 2 so the solution will be c1 take exponential this then the power is 4 times x this is the way to write the solution of this equation plus c2 exponential minus 2 x so this is the required solution of your differential equation so from here we can prove we check that this is the solution of differential equation so one thing is for so for second equation gx is equal to let me write gx same as first only that differential equation is different c1 e2 x c2 x c1 exponential 2x plus c2 x2x plus c2 x2x plus c2 xxxxxxxx22x plus c2 x2xx plus c2 x2x plus c2 xx plus c3 into exponential minus 2x.

04:18 C1, c2, c3 are the constants.

04:21 Is the solution of differential equation is d cube y by dx cube minus 2, d2y over dx square minus 4, dy over dx plus 8y is equal to 0.

04:49 So, in this case also i'm using the normal equation to solve this equation.

04:55 Just convert into the d form...

Question

There are functions of the form $e^{rx}$ that solve the differential equation Identify a fundamental set of solutions for this differential equation. $\{e^{-2x}, e^{4x}\}$ $\{e^{-4x}, e^{-2x}\}$ $\{xe^{-2x}, e^{-2x}\}$ $\{e^{-4x}, e^{2x}\}$ $\{e^{2x}, e^{4x}\}$ $y'' + 6y' + 8y = 0$

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